A man has $80\text{kg-wt}$ on the surface of Earth of radius $\left(R\right)$. At what height above the surface of the Earth his weight will be $40\text{kg-wt}$ ?

The correct option is C $(\sqrt{2}-1)R$
Weight of body on the earth's surface is $w=80\text{Kg-wt}$

So, $w=mg$ and, Weight at a height h above the earth's surface $w^{\prime }=40\text{Kg-wt}$

So, $w^{\prime }=\frac{mg}{{(1+\frac{h}{R})}^2}$

$\therefore \frac{w}{w^{\prime }}={(1+\frac{h}{R})}^2$

or, $2=(1+h/R{)}^2$

$\sqrt{2}=1+\frac{h}{R}$

$\frac{h}{R}=\sqrt{2}-1$

$h=(\sqrt{2}-1)R$